On Fri, Oct 3, 2014 at 9:22 PM, drago01 <drago01@xxxxxxxxx> wrote:
> but given that F<n> => <Fn+1> is tested
> and supposed to work anyway
> this should "just work" without any additional effort.
There is some room for breakage between
F<n> at the time F<n+1> is released => F<n+1> at the time F<n+1> is released
and
F<n> at the time F<n+4> is released => F<n+1> at the time F<n+4> is released,
but hopefully not too much
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